What's exactly the new definition of kilogram, second and meter? Could one explain  this?

Technically a kilogram (kg) is now defined:
[…] by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10–34 when expressed in the unit J s, which is equal to kg m2 s–1, where the metre and the second are defined in terms of c and ΔνCs.

Does that mean that 1 kg = 1 Planck constant?
And what exactly is the new definition of the second and the meter?
 A: The definitions of "second" and "metre" have not changed.
One second is defined in terms of frequency. Frequency is measured in hertz $(1\ \rm Hz=1\ s^{-1}$
We take an atom of $\rm Cs$. And then, we count the frequency of its spectrum. We extract the unit "1 second" from there.
As for the meter, we set that "one metre is the distance light travels in $\frac{1}{2,997955}~\rm s".$
So the meter and the second are perfectly defined.

The new thing is that the kilogram is no longer "the mass of a weight located in Paris, France". Now we have redefined it in terms of absolute things.
If you take the actual definition of metre and second, Plank's constant is
$$h=6,626\ldots \times 10^{34}~\rm Js$$
with many decimal numbers.
So we say "okay, let's cut the decimals somewehre". Let's say that Plank's constant is now EXACTLY
$$h:= 6.626 070 15 \times 10^{-34}~\rm Js$$
And then we say "adapt the value of $1~\rm kg$ so that Plank's constant is exactly that one.
A: The SI system is now defined entirely by physical constants. There are no more “prototype” artifacts. How it works is thus:
https://www.bipm.org/utils/common/pdf/si-brochure/SI-Brochure-9-concise-EN.pdf

The SI is the system of units in which:


• the unperturbed ground state hyperfine transition frequency of the caesium 133 atom $\Delta \nu_{Cs}$ is $9 \ 192 \ 631 \ 770 \text{ Hz}$,


• the speed of light in vacuum $c$ is $299 \ 792 \ 458 \text{ m/s}$,


• the Planck constant $h$ is $6.626 \ 070 \ 15 × 10^{−34} \text{ J s}$,

So we make a caesium atomic clock and when that clock ticks 9192631770 times, that is $1\text{ s}$. This is our SI unit of time.
Now, we want to get a unit of distance. We could use a physical prototype but physical prototypes can be damaged or distorted and are by definition not possible to distribute.
So instead we can use a universal physical constant, in this case the speed of light. We simply define the meter to be the length such that the speed of light in a vacuum is exactly $299792458\text{ m/s}$. Since we already have a definition for a second, fixing the speed of light defines the meter. Any experiment that we previously would have used to measure the speed of light now becomes a measurement of the length of a meter. This is good because we can measure the speed of light very accurately and it allows a meter that cannot be damaged or distorted and which anyone can replicate.
Then we take the same approach for mass. If we define Planck’s constant to be exactly $6.62607015 × 10^{−34} {\rm\ kg\ m^2\ s^{-1}}$ then, since we already have fixed the second and the meter this defines the kilogram. Any experiment that we previously would have used to measure Planck’s constant now becomes a measurement of the mass of a kilogram. It has all of advantages of the meter, except currently Planck constant experiments are not as precise as speed of light experiments.
A: The other answers are correct, but I don't think they've been very direct in addressing your specific question,

Does that mean that 1 kg = 1 Planck constant?

Planck's constant $h$ is (by definition) $ 6.626 070 15 × 10^{–34}{\rm\ J\ s}$.
Since $1\ {\rm J}=1\ {\rm kg\ m\ s^{-2}}$, we have
$$1\ {\rm kg} = \frac{h}{6.62607015\times 10^{-34}\ {\rm m\ s^{-1}}}$$
